What is the integral from 0 to 1 of x/(sqrt(1+3x^2)) ?

The integral from 0 to 1 of x/(sqrt(1+3x^2)) is 1/3

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integral from 0 to 1 of x/(sqrt(1+3x^2))

\int_{0}^{1} \frac{x}{1 + 3 x ^{2}} d x

\frac{1}{3}

Solution steps

\int_{0}^{1} \frac{x}{1 + 3 x ^{2}} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{6 u} d u

= \int_{1}^{4} \frac{1}{6 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{1}^{4} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{4}

= \frac{1}{6} [2 u^{\frac{1}{2}}]_{1}^{4}

Apply radical rule:

= \frac{1}{6} [2 u]_{1}^{4}

Compute the boundaries:

2

= \frac{1}{6} \cdot 2

Simplify

= \frac{1}{3}

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What is the integral from 0 to 1 of x/(sqrt(1+3x^2)) ?
The integral from 0 to 1 of x/(sqrt(1+3x^2)) is 1/3